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Stoichiometry: Real World Reactions

Energy Changes

Problems

Problems

Exothermic and Endothermic Reactions

Some stoichiometric calculations involve the change in energy that accompanies a chemical reaction. Reactions that release energy in the form of heat are called exothermic reactions. Conversely, reactions requiring heat energy are known as endothermic reactions. Here are some examples of exothermic and endothermic reactions:

Exothermic:


    C(s) + O2(g)→CO2(g) + 393.5 kJ  
    SO3(g) + H2O(l)→H2SO4(aq) + 129.6 kJ  

Endothermic:


    CaCO3(s) + 176 kJ→CaO(s) + CO2(g)  
    2CO2(g) + 43.9 kJ→2CO(g) + O2(g)  

Reactions like this, which include the amount of heat energy produced or absorbed, are known as thermochemical equations. Although not common, note that these equations can also be written as follows:

Exothermic:


    C(s) + O2(g) - 393.5 kJ→CO2(g)  
    SO3(g) + H2O(l) - 129.6 kJ→H2SO4(aq)  

Endothermic:


CaCO3(s) CaO(s) + CO2(g) - 176 kJ2CO2(g)  
  2CO(g) + O2(g) - 43.9 kJ  

Heat Energy and Mole Ratios

When calculating mole ratios for reactions with given heat energies, you must treat the heat energy just as you would any other reactant/product in the reaction. How? Just imagine there is a coefficient of 1 in front of the heat energy. This amount of heat energy is absorbed/produced in ratio with the rest of the equation. This concept is illustrated in the following example problem.

Problem: Based on the following balanced equation, how many kcal of energy are needed to decompose 200 grams of CaCO3(s) ?

CaCO3(s) + 176 kJ→CaO(s) + CO2(g)    

Solution: Convert to moles.

×1 mole CaCO3 = 3 moles CaCO3    

Now do the mole ratio; 176 kJ are needed for every mole of CaCO3 .

×176 kJ = 528 kJ    

Finally, you must put the answer in terms of kcal. There are 4.18 kJ in every kcal. Use the necessary conversion factor.

= 126 kcal    

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